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What is the derivative of #f(x) = ln(sin^2x)#?

Answer 1

Samantha Adkison

Applying the chain rule, #d/dxln(sin^2(x)) = 2cot(x)#

The chain rule states that #d/dxf(g(x)) = f'(g(x))*g'(x)#

Let #f(x) = ln(x)# and #g(x) = sin^2(x)#

Then #f'(x) = 1/x#

To find #g'(x)# we need to use the chain rule again with #g_1(x) = x^2# and #g_2(x) = sin(x)#

Then #g_1'(x) = 2x# and #g_2'(x) = cos(x)# So, as #g(x) = g_1(g_2(x))# #g'(x) = g_1'(g_2(x))*g_2'(x) = 2sin(x)*cos(x)#

Going back to the original problem, we have #ln(sin^2(x)) = f(g(x))#

so, applying the chain rule, we get #d/dxln(sin^2(x)) = f'(g(x))*g'(x) = 1/(sin^2(x))*2sin(x)cos(x)#

Ultimately, simplifying yields the ultimate outcome of

#d/dxln(sin^2(x)) = 2cot(x)#

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Answer 2

Isabella Ackerman

The derivative of ( f(x) = \ln(\sin^2x) ) is ( \frac{d}{dx}\left(\ln(\sin^2x)\right) = \frac{1}{\sin^2x} \cdot \frac{d}{dx}(\sin^2x) ). Using the chain rule, ( \frac{d}{dx}(\sin^2x) = 2\sin(x) \cdot \cos(x) ). Thus, the derivative of ( f(x) ) is ( \frac{1}{\sin^2x} \cdot 2\sin(x) \cdot \cos(x) = \frac{2\sin(x) \cdot \cos(x)}{\sin^2x} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.